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is the second equation \[y=2x-3\]?
Well, let's assume that it is. \[y = (x+1)^2-2\]\[y=2x-3\] We can solve this by substitution. Everywhere you see \(y\) in the first equation, replace it with the right hand side of the second equation (which is also equal to \(y\)). This gives us an equation all in terms of just the variable \(x\): \[2x-3 = (x+1)^2-2\]Now you can expand that out: \[2x-3 = (x+1)(x+1)-2\]\[2x-3=x^2+1x+1x+1*1-2\]\[2x-3=x^2+2x-1\]and get all the terms on one side and you have a quadratic to solve. You know how to do that?
The second equation is y-2x=3
Okay, can you solve \[y-2x=3\] for \(y\)?
Right. So now we do the same thing I did, except with the correct equation. \[y = (x+1)^2+2\]\[y = 2x+3\] \[2x+3 = (x+1)^2 + 2\] Expand both sides, then combine all the terms on one side and solve the quadratic.
wait it should be 3+2x=(x+1)^2-2
doesn't \[2x+3 = 3+2x\]?
sorry, I switched a sign, that's what you were pointing out, my bad!
\[3+2x = (x+1)^2-2\]
yes but the original equation is y=(x+1)^2-2 and if we are plugging in y , and y=2x+3
Yeah thats what I got :))
Glad one of us is paying close attention :-)
so now i get 2x+3=x^2+1+2
thats after expanding the equation
close... \[(x+1)^2 = (x+1)(x+1) = x*x + 1*x + 1*x + 1*1 = x^2 + 2x + 1\] so [2x+3 = x^2+2x+1 - 2\]\[2x+3 = x^2+2x-1\] right?
yes got that
okay, get all the terms collected on one side, what do you get?
Okay, and what are the solutions to that equation?
That's a simple one to solve: \[x^2-4 = 0\]\[x^2 = 4\]Take the square root of both sides and you see \[x=\pm 2\] Have to be sure to include the negative square root or we won't get all of the answers! Now we need to find the values of \(y\) that go along with those values of \(x\). We can use either equation, so might as well use the simpler one: \[y=2x+3\] Plug in both values of \(x\) and get the corresponding values of \(y\).
okay just a sec
I get (2,7) and (-2,-1)
So do I!
and we can easily test them: \[2(2)+3 = (2+1)^2-2\]\[7 = 9-2\checkmark\] \[2(-2)+3 = (-2+1)^2-2\]\[-1=1-2\checkmark\]
So we can do this any time we can rearrange the stuff in one equation to give us something we can substitute into the other one.
Here's a tip that doesn't really matter here so much, but does in problems that you will encounter: always try your answers out in the original equation and make sure they work! Whenever you have to square both sides of the equation, you can get what called "spurious" solutions which come out just the same way as the real ones do, but don't actually make the equations true when you plug the numbers in. Rather than remembering exactly what kinds of problems require you to do this, I prefer to simply always check my answers!